Left Termination of the query pattern app_in_3(g, g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

app([], X, X).
app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).

Queries:

app(g,g,a).

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
app_in(x1, x2, x3)  =  app_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
[]  =  []
app_out(x1, x2, x3)  =  app_out(x3)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
app_in(x1, x2, x3)  =  app_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
[]  =  []
app_out(x1, x2, x3)  =  app_out(x3)


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

APP_IN(.(X, Xs), Ys, .(X, Zs)) → U11(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
app_in(x1, x2, x3)  =  app_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
[]  =  []
app_out(x1, x2, x3)  =  app_out(x3)
U11(x1, x2, x3, x4, x5)  =  U11(x1, x5)
APP_IN(x1, x2, x3)  =  APP_IN(x1, x2)

We have to consider all (P,R,Pi)-chains

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof

Pi DP problem:
The TRS P consists of the following rules:

APP_IN(.(X, Xs), Ys, .(X, Zs)) → U11(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
app_in(x1, x2, x3)  =  app_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
[]  =  []
app_out(x1, x2, x3)  =  app_out(x3)
U11(x1, x2, x3, x4, x5)  =  U11(x1, x5)
APP_IN(x1, x2, x3)  =  APP_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 1 less node.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof

Pi DP problem:
The TRS P consists of the following rules:

APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
app_in(x1, x2, x3)  =  app_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
[]  =  []
app_out(x1, x2, x3)  =  app_out(x3)
APP_IN(x1, x2, x3)  =  APP_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

Pi DP problem:
The TRS P consists of the following rules:

APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
APP_IN(x1, x2, x3)  =  APP_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ QDPSizeChangeProof

Q DP problem:
The TRS P consists of the following rules:

APP_IN(.(X, Xs), Ys) → APP_IN(Xs, Ys)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: